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Prove that $$\frac{\text{lcm}(x,y)}{\text{gcd}(x,y)}\in\Bbb N$$ for all $x, y \in \Bbb N$. And what kind of conditions you need, so $\frac{\text{lcm}(x,y)}{\text{gcd}(x,y)}$ would be equal integer's square?

AHB
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user401272
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    We can follow the pattern from http://math.stackexchange.com/questions/2120744/gcd-and-lcm-theorem/2120753#2120753 – lab bhattacharjee Jan 30 '17 at 12:18
  • I think the solution becomes obvious once you see that $LCM(x,y)=p_1^{\max(x_1,y_1)}p_2^{\max (x_2,y_2)}...p_n^{\max(x_n,y_n)}$ and $GCD(x,y)=p_1^{\min(x_1,y_1)}p_1^{\min(x_n,y_n)}...p_n^{\min(x_n,y_n)}$ where $p_1^{x_1}p_2^{x_2}..p_n^{x_n}$ is the prime factorization of x and $p_1^{y_1}p_2^{y_2}..p_n^{y_n}$ is the prime factorization of y – Andreas Ch. Jan 30 '17 at 12:39
  • Do you realize this question is the same thing as asking you to prove $\gcd(x,y) \mid \operatorname{lcm}(x,y)$? –  Jan 30 '17 at 12:40

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Hint $\,\ d\mid x\mid m\,\Rightarrow\, d\mid m\,$ by transitivity of divisibility.

As for the second question, note that a positive integer is a square iff every prime occurs to even power in its unique prime factorization. But the power of a prime $p$ in the lcm/gcd is even iff its power in $x$ and $y$ have equal parity (since it's their max $-$ min in lcm/gcd). This is equivalent to $x/y$ being a square of a rational, or $xy$ being a square.

Bill Dubuque
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  • i.e. by gcd & lcm universal property $,\gcd(a,b)\mid a\mid {\rm lcm}(a,b),$ so the claim follows by transitivity of divisibility.

    It boils down to $,\inf S \le \sup S,$ in poset / lattice language. More generally, the hint shows that every common divisor of a set of integers divides every common multiple.

     – Bill Dubuque Aug 13 '21 at 16:50